First the verdict on the posted problem. It has certainly no solution in up to 12 moves (checked by Popeye v4.79). And the author of the Gustav software which specializes in selfmates says that there is probably no regular solution to this problem.
Another expert thinks that if White simply captures everything which moves, he can then push bK over to g3 and force Black to win by h3xg2#. That's like taking a long chess problem and solving it like an regular chess game with no limit to the number of moves. Yes it will work to win but it's not really in the spirit of things. That's not to diminish his ingenious approach, but it's just not what we would consider a solution.
The good news is that this problem turns out to be a recognizable example of the Broecker schema, invented in 1891 by Gustav Von Broecker. The German PDB database contains 36(!) other problems which are all based on the same diagonal opposition of bishops. One early very clear example is the following:
[Title "Broecker - London Chess Fortnightly 1892 - s#9"]
[fen "kb4R1/1b6/P7/8/8/7p/6BP/6BK w - - 0 0"]
1. Rf8 Bc6 2. Re8 Bd5 3. Rd8 Be4 4. Rc8 Bf3 5. Rh8 Be4 6. Bf3 Bd5 7. Be4 Bc6 8. Bd5 Bb7 9. Bc6 Bxc6#
"s#9" is shorthand for "selfmate in 9" - i.e. White moves first and forces an uncooperative Black to checkmate White by Black's 9th move at the latest.
As the original poster had deduced, there is a key relationship between the square of the queen (or rook) on the 8th rank, with the number of squares separating the bishops. Either player must always make them match if they can, and if he fails to, the other player gets a chance to make them match instead.
White wins because he is able to match by moving 1. Tf8 only. Black has many possible lines of play, but the longest he can hold out is 9 moves. One example solution is shown here.
In the original problem, the position is already a "match" with White to play. He is forced to break the match, and Black can always recreate it, so need never lose. It doesn't seem there is any way for White to use the additional power of the queen to change the parity.
However the matrix has proved very rich over the years, as you can see from the database, and I am sure there's more that can still be mined. The longest problem known to be sound is the following:
[Title "Rotenburg & Trillon - Die Schwalbe 1980 - s#14"]
[fen "k7/1b4R1/N7/7p/8/7p/6BP/6BK w - - 0 0"]
1. Rf7! Bc6 2. Re7 Bd5 3. Rd7 Be4 4. Rc7 Bf3 5. Rh7 Be4 6. Bf3 Bd5 7. Be4 Bc6 8. Bd5 Bb7 9. Bc6 h4! 10. Bd5 Bc6 11. Rc7 Bb7 12. Rc1 Bc6 13. Rb1 Bb7 14. Bc6 Bxc6#
Again, here is a sample solution as Black has multiple choices, although White's move is unique at every point.
What's interesting about this one is that Black has a waiting move. This flips the parity, and so normally Black would be able to continue making matches, while White is breaking matches. However, White can flip the parity, by moving to the other side of the main diagonal. This can only be done while ensuring that b7 is never free for bK to enter. Either it must be occupied by bB, or attacked by wR.